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80
229
19.6
5
z p
y
z i
y
612.2
80
1 1 . 9
10
z
z
(a) 610 × 229 UB 125
I y = 98 610 cm 4
W pl,y = 3676 cm 3
(b) Tee-section
Figure 5.40 Worked examples 1, 2, and 18.
Solution . Using the thin-walled assumption and equations 5.57 and 5.58,
z i = ( 80 + 5 / 2 ) × 10 × ( 80 + 5 / 2 )/ 2
( 80 × 5 ) + ( 80 + 5 / 2 ) × 10
= 27.8 mm
I y = ( 80 × 5 ) × 27.8 2 + ( 82.5 3 × 10 / 12 ) + ( 82.5 × 10 )
× ( 82.5 / 2 27.8 ) 2 mm 4
= 92.63 cm 4
W el , y = 92.63 / { ( 8.25 2.78 )/ 10 }= 16.93 cm 3
Alternatively, the equations of Figure 5.6 may be used.
5.12.3 Example3-properties of an angle section
Problem . Determinetheproperties A , I y 1 , I z 1 , I y 1 z 1 , I y , I z ,and α ofthethin-walled
angle section shown in Figure 5.41c.
Solution . Using the thin-walled assumption, the angle section is replaced by two
rectangles 142.5 ( = 150 15 / 2 ) × 15 and 82.5 ( = 90 15 / 2 ) × 15, as shown in
 
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